leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

Paradoxica · Feb 3, 2016

leehuan said:
$\\ $This question is more or less out of interest. \\ \\ Can dot products in 2 dimensions be related in any way whatsoever to Ext 2 complex numbers?$

Given the fact that both of them require two dimensional mathematical reasoning, yes. Anything interpreted vectorially can be re-interpreted geometrically, which is identical to the complex field.

drsabz101 · Feb 4, 2016

Help ASAP

Solve these 3 equations to find real number p,q,r

P+2q-r=5
2p-q+r=3
P+3q-2r=8

InteGrand · Feb 4, 2016

drsabz101 said:
Help ASAP

Solve these 3 equations to find real number p,q,r

P+2q-r=5
2p-q+r=3
P+3q-2r=8

$\noindent We have the system$

$\left\{\begin{matrix} p+2q-r=5& (1) & \\2p-q+r=3 & (2)& \\ p+3q-2r = 8& (3) & \end{matrix}\right.$

$\noindent The goal is to first eliminate one of the variables completely, so we are left with only two variables.$

$(1) + (2): 3p + q = 8\quad (4)$

$2\times (2)+ (3): 5p +q = 14 \quad (5)$

$\noindent This is a system of equations with two unknowns and two equations, which you should know how to go about. From (4), we have $q = -3p + 8$. Now, sub. this into (5): $5p +(-3p+8)=14 \Rightarrow 2p = 6 \Rightarrow p = 3$.$

$\noindent So $q = -3p + 8\Rightarrow q = -3\cdot 3 + 8=-1$. Now, from (2), $r= 3-2p + q = 3 -2\cdot 3 +\left(-1\right)\Rightarrow r=-4$. So the solution is $(p,q,r) = \left(3,-1,-4\right)$.$

Speed6 · Feb 4, 2016

drsabz101 said:
Help ASAP

Solve these 3 equations to find real number p,q,r

P+2q-r=5
2p-q+r=3
P+3q-2r=8

Correct me if I am wrong, but it isn't this just simultaneous equations?

I.e make a subject and sub in and then evaluate etc with respect to all 3 variables?

InteGrand · Feb 4, 2016

Speed6 said:
Correct me if I am wrong, but it isn't this just simultaneous equations?

Yeah, this is just simultaneous equations.

drsabz101 · Feb 5, 2016

how do I find the natural domain of :

InteGrand · Feb 5, 2016

drsabz101 said:
how do I find the natural domain of :View attachment 32856

$\noindent For the first one, we want $x^2 \leq 9\Longleftrightarrow -3\leq x \leq 3$. So the domain is $\{ x\in \mathbb{R}: -3\leq x\leq 3\}$.$

$\noindent For the second, the domain is just $\mathbb{R}^+$ (the set of all positive reals).$

$\noindent For the third, the domain is all real $x$ excluding $-1$, i.e. $\mathbb{R} \backslash \{-1\}$.$

$\noindent For the fourth, we need $x^2-4 \geq 0\Longleftrightarrow |x| \geq 2$. So the domain is $\{x\in \mathbb{R}: |x|\geq 2\}$.$

drsabz101 · Feb 5, 2016

Thankyou

drsabz101 · Feb 5, 2016

I have also other questions uploaded here: View attachment 32857 and

thanks again

drsabz101 · Feb 5, 2016

Also need help solving and sketching this inequality :

authenticity · Feb 5, 2016

Here's the answer for the Qs in the FullSizeRender image http://imgur.com/a/9RDEl . Attachment 32857 leads to no where.

drsabz101 · Feb 5, 2016

thanks

authenticity · Feb 5, 2016

You're welcome! Here's the link for the next anwers: http://i.imgur.com/8CAEdur.jpg?1

Hopefully they're right. I haven't done maths harder than addition since HSC LOL

authenticity · Feb 5, 2016

Oh wait, I shaded the second one wrong. It's supposed to be http://i.imgur.com/QRL32yb.jpg?1

leehuan · Feb 5, 2016

InteGrand · Feb 5, 2016

leehuan said:
$\\ $To reduce the reading I have to do, can someone please explain the deal with this?$ \\ \int { \frac { dx }{ 1-{ x }^{ 2 } } } \\ = \begin{cases} \tanh ^{ -1 }{ x }, \left| x \right| <1 \\ \coth ^{ -1 }{ x }, \left| x \right| >1 \end{cases}$

$\noindent Using partial fractions, the primitive is $\frac{1}{2}\ln \left| \frac{1+x}{1-x}\right|$, $x \neq \pm 1$ (if $x=\pm 1$, the argument of the log is either undefined or 0, neither of which is permissible).$

$\noindent So your statement is essentially just saying that for $|x|<1$, this log thing is the expression for the inverse hyperbolic tangent, and for $|x|>1$, it is the expression for the inverse hyperbolic cotangent (well, technically these could differ from that by a constant of integration, but they don't in fact). These can be proved by using the definitions of $\tanh$ and $\coth$ (in terms of exponentials) and forming a quadratic to solve for the inverse. Also note that the domain of $\tanh ^{-1} x$ is $\{x\in \mathbb{R}:|x|<1\}$ (since the range of $\tanh$ is $(-1,1)$), and the domain of $\coth^{-1}$ is $\{ x\in \mathbb{R}: |x|>1\}$ due to a similar reason (range of $\coth$ being $\{ y\in \mathbb{R}: |y|>1\}$). This explains why which function appears as the primitive depending on whether $|x|>1$ or $<1$.$

drsabz101 · Feb 6, 2016

I am really confused on how to draw inverse functions . Can someone please help. I just don't understand where I am reflecting the graphs.

Shuuya · Feb 6, 2016

drsabz101 said:
I am really confused on how to draw inverse functions . Can someone please help. I just don't understand where I am reflecting the graphs.

You reflect the graph about the line y=x, which is the dotted line in the diagram below

Paradoxica · Feb 6, 2016

Can someone explain to me why there are no higher order analogues to the Gaussian Integers and Eisenstein Integers?

And for the sake of normal people, you are allowed to provide geometric intuition.

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